JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Elasticity & SHM

Two slabs with square cross section of different materials

(1, 2)

with equal sides

(l)

and thickness

d_{1}

and

d_{2}

such that

d_{2} = 2 d_{1}

and

l > d_{2}

. Considering lower edges of these slabs are fixed to the floor, we apply equal shearing force on the narrow faces. The angle of deformation is

θ_{2} = 2 θ_{1}

. If the shear moduli of material 1 is

4 \times 1 0^{9} N / m^{2}

, then shear moduli of material 2 is

x \times 1 0^{9} N / m^{2}

, where value of

x

is ______.

Answer: 1

Solution

For shear, the angle of deformation equals the shear stress divided by the shear modulus,

θ = F / (A η)

, where the force is spread over the narrow face of area

A \propto l d

. With the same force and the same width

l

θ_{1} = \frac{F}{l d _{1} η _{1}}, θ_{2} = \frac{F}{l d _{2} η _{2}} .

Imposing

θ_{2} = 2 θ_{1}

with

d_{2} = 2 d_{1}

\frac{F}{l ( 2 d _{1} ) η _{2}} = 2 \cdot \frac{F}{l d _{1} η _{1}} \Rightarrow η_{2} = \frac{η _{1}}{4} = \frac{4 \times 1 0 ^{9}}{4} = 1 \times 1 0^{9} N/m^{2} .

The value of

x

1

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Integer answer